Let $V$ be a real inner product space, $\{x_{n}\mid n\in\mathbb{N}\}$ is an orthonormal basis.
I want to show that $\phi : V \rightarrow l_2\;$ which takes each $v\in V$ to $\left\langle v,x_{n}\right\rangle _{n=1}^{\infty}\in l_{2}$, is distance preserving.
$l_{2}=\{(x_{n})_{n=1}^{\infty}\mid\sum_{n=1}^{\infty}\left|x_{n}\right|^{2}<\infty\}$
I tried by using some indenties and polarization but i'm stuck.
I know that:
$d_{V}(u,v)=\left\langle u,v\right\rangle =_{polarization}\frac{1}{2}(\left\Vert u+v\right\Vert ^{2}-\left\Vert u\right\Vert ^{2}-\left\Vert v\right\Vert ^{2})=_{Parallelogram\;law}\frac{1}{2}(2(\left\Vert u\right\Vert ^{2}+\left\Vert v\right\Vert ^{2}-\left\Vert u-v\right\Vert ^{2})-\left\Vert u\right\Vert ^{2}-\left\Vert v\right\Vert ^{2})$
I need to show that $d_{V}(u,v)=d_{l_{2}}(\phi(u),\phi(v))\;$ for all $u,v \in V$
Let $v\in V$, since $x_n$ is an orthogonal basis, write $v=\sum v_nx_n$, we have $\langle v,x_n\rangle=v_n$. Let $w=\sum w_nx_n$ $d(v,w)^2=\sum |v_n-w_n|^2$
We have $\phi(v)=(,..v_n,..)$. , we have $\phi(w)=(,..,w_n,..)$ This implies that implies that $|\phi(v)-\phi(w)|^2=\sum|v_n-w_n|^2$.